Question Video: Identifiying Convergent and Divergen Natural Logarithm Series Using the Integral Test Mathematics • Higher Education

Use the integral test to determine whether the series βˆ‘_(𝑛 = 1) ^∞ (ln (𝑛))/(𝑛) is convergent or divergent.

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Video Transcript

Use the integral test to determine whether the series the sum from 𝑛 equals one to ∞ of the natural logarithm of 𝑛 divided by 𝑛 is convergent or divergent.

The question wants us to determine the convergence or divergence of a series by using the integral test. So let’s start by recalling what the integral test tells us. The integral test says if we have a function 𝑓 of π‘₯ which is positive, continuous, and decreasing for all values of π‘₯ greater than or equal to some constant π‘˜. And if we have a sequence π‘Ž 𝑛 which is equal to 𝑓 evaluated at 𝑛 for all of our values of 𝑛. Then the integral test tells us if the integral from π‘˜ to ∞ of 𝑓 of π‘₯ with respect to π‘₯ is convergent, then our sum from 𝑛 equals π‘˜ to ∞ of π‘Ž 𝑛 must also be convergent. However, if the integral from π‘˜ to ∞ of 𝑓 of π‘₯ with respect to π‘₯ is divergent, then our sum from 𝑛 equals π‘˜ to ∞ of π‘Ž 𝑛 must also be divergent.

The integral test gives us a method of turning a question about the convergence or divergence of an infinite series into a question about the convergence or divergence of an integral. Since the question wants us to use the integral test on the sum from 𝑛 equals one to ∞ of the natural logarithm of 𝑛 divided by 𝑛, we’ll set our value of π‘˜ equal to one. And our function 𝑓 of π‘₯ to be equal to the natural logarithm of π‘₯ divided by π‘₯. Now, to use the integral test, we’re going to need to show that our function 𝑓 of π‘₯ which is equal to the natural logarithm of π‘₯ divided by π‘₯ is a positive, continuous, and decreasing function for all values of π‘₯ greater than or equal to one.

If we were to try and show that our function 𝑓 of π‘₯ was positive on this interval, we would immediately run into a problem. We see that when π‘₯ is equal to one, our function is equal to the natural logarithm of one divided by one. However, the natural logarithm of one is equal to zero. So we see that 𝑓 of one is equal to zero. So our function 𝑓 of π‘₯ is not positive on this interval. However, we can easily get around this. We notice this is also the first term in our series. If the first term in our series is equal to zero, then we can just take this outside of our series. It won’t change the value of our infinite sum.

In other words, the sum from 𝑛 equals one to ∞ of the natural logarithm of 𝑛 divided by 𝑛 is equal to the sum from 𝑛 equals two to ∞ of the natural logarithm of 𝑛 divided by 𝑛. So we’ll instead try to use the integral test on this second series. We’ll set our value of π‘˜ equal to two. And we leave our function 𝑓 of π‘₯ to be the natural logarithm of π‘₯ divided by π‘₯. So let’s clear some space and update our integral test for our value of π‘˜ to be equal to two. We now need to show that our function 𝑓 of π‘₯ is positive, continuous, and decreasing for all values of π‘₯ greater than or equal to two.

So let’s start by showing our function 𝑓 of π‘₯ is positive on this interval. If π‘₯ is greater than or equal to two, we know the natural logarithm of π‘₯ is positive and π‘₯ is positive. So this means when π‘₯ is greater than or equal to two, our function 𝑓 of π‘₯ is the quotient of two positive numbers. This means our function 𝑓 of π‘₯ is positive for π‘₯ is greater than or equal to two. We now want to check our function 𝑓 of π‘₯ is continuous for values of π‘₯ greater than or equal to two. Well, we know the numerator, the natural logarithm of π‘₯, is continuous and the denominator of π‘₯ is also continuous. So 𝑓 of π‘₯ is the quotient of two continuous functions. And we know the quotient of two continuous functions will be continuous across its entire domain.

So we need to check the domain of our function 𝑓 of π‘₯. We can’t divide by zero, so we know π‘₯ is not allowed to be equal to zero. And we can’t take the logarithm of a number less than or equal to zero. So the domain of our function 𝑓 is all values of π‘₯ greater than zero. In particular, that means that 𝑓 of π‘₯ is continuous for values of π‘₯ greater than or equal to two. All we need to show now is that our function 𝑓 of π‘₯ is decreasing for all values of π‘₯ greater than or equal to two. We’ll do this by finding the slope of our function. To do this, we’ll start by considering our function 𝑓 of π‘₯ to be the product of two functions. It’s the product of the natural logarithm of π‘₯ and one over π‘₯.

We can then differentiate this by using the product rule. Remember, the product rule tells us the derivative of 𝑒 of π‘₯ times 𝑣 of π‘₯ with respect to π‘₯ is equal to 𝑒 prime of π‘₯ times 𝑣 of π‘₯ plus 𝑣 prime of π‘₯ times 𝑒 of π‘₯. So we can find an expression for 𝑓 prime of π‘₯. We know the derivative of the natural logarithm of π‘₯ with respect to π‘₯ is one over π‘₯. And we know the derivative of the reciprocal function with respect to π‘₯ is negative one divided by π‘₯ squared. Using this in the product rule, we get 𝑓 prime of π‘₯ is equal to one over π‘₯ times one over π‘₯ plus negative one over π‘₯ squared times the natural logarithm of π‘₯. And then, we’ll simplify this expression to get one minus the natural logarithm of π‘₯ all divided by π‘₯ squared.

Remember that this represents the slope of our function 𝑓 of π‘₯. And we know that our function will be decreasing when its slope is negative. So to find the values of π‘₯ where our function is decreasing, we just need to find the values of π‘₯ where this expression is negative. We can see the denominator of our function is π‘₯ squared. Since we’re only interested in π‘₯ being greater than or equal to two, we can see that this will always be positive. So the only way for our slope to be negative is when the numerator is negative. The numerator is negative when one minus the natural logarithm of π‘₯ is less than zero or, alternatively, one is less than the natural logarithm of π‘₯.

And we know how to solve this. We know when π‘₯ is equal to 𝑒, the natural logarithm of π‘₯ will be equal to one. We also know the natural logarithm function is an increasing function. So when π‘₯ is greater than 𝑒, the natural logarithm of π‘₯ will be greater than one. So what have we actually shown? We’ve shown the slope of our function 𝑓 of π‘₯ is negative when π‘₯ is greater than 𝑒. So this means our function 𝑓 of π‘₯ is decreasing for all values of π‘₯ greater than or equal to 𝑒. But this is not enough to use our integral test in its current form. We needed our function to be decreasing for all values of π‘₯ greater than or equal to two.

However, we know our value of 𝑒 is approximately 2.7. So we have in fact shown that our function 𝑓 of π‘₯ is decreasing for all values of π‘₯ greater than or equal to three. So, in fact, we could try using the integral test with our value of π‘˜ equal to three. And this would still give us an answer to our original question. Remember, the sum from 𝑛 equals one to ∞ of the natural logarithm of 𝑛 divided by 𝑛 is equal to the sum from 𝑛 equals three to ∞ of the natural logarithm of 𝑛 divided by 𝑛. Where we add the first and second term of our sequence. So the convergence or divergence of these two series will be the same.

So let’s rewrite our integral test this time with π‘˜ equal to three. And we keep 𝑓 of π‘₯ to be the natural logarithm of π‘₯ divided by π‘₯. Well, we’ve already shown that this function is positive for these values of π‘₯. We’ve already shown it’s continuous for these values of π‘₯. And we’ve already shown it’s decreasing for these values of π‘₯. So now that we’ve shown all of our prerequisites for the integral test are true, let’s clear some space and work on applying the integral test.

First, we might be worried about the concluding statement of the integral test. We now have a conclusion about the convergence or divergence of the sum from 𝑛 equals three to ∞ of our sequence π‘Ž 𝑛. Well, in our case, our sequence π‘Ž 𝑛 is the natural logarithm of 𝑛 divided by 𝑛. And we know the sum from 𝑛 equals one to ∞ of the natural logarithm of 𝑛 divided by 𝑛 is equal to the first term plus the second term plus the sum from 𝑛 equals three to ∞ of the natural logarithm of 𝑛 divided by 𝑛. And both the natural logarithm of one divided by one and the natural logarithm of two divided by two are constants. So these two series differ by some constant. In fact, we know they differ by the natural logarithm of two divided by two. So this tells us the convergence or divergence of both of these two series are the same.

So this justifies us using the integral test in this way to determine the convergence or divergence of the series given to us in the question. Now, to use the integral test, we need to determine the convergence or divergence of the integral from three to ∞ of the natural logarithm of π‘₯ divided by π‘₯ with respect to π‘₯. First, we’ve already shown that our integrand, the natural logarithm of π‘₯ divided by π‘₯, is continuous on our interval of integration. It’s continuous for all values of π‘₯ greater than or equal to three.

So the only thing about this integral which is improper is the upper limit being ∞. So by using our rules for improper integrals, we can write this integral as the limit as 𝑑 approaches ∞ of the integral from three to 𝑑 of the natural logarithm of π‘₯ divided by π‘₯ with respect to π‘₯. And if this limit converges, we say our integral converges. However, if this limit diverges, then we say our integral diverges. And it’s worth pointing out now the upper limit of our integral is some finite positive number 𝑑. So our integrand is continuous on the entire interval of integration. So we can use any of our tools to evaluate this integral.

There’s a few different ways to evaluate this integral. The easiest is to notice that the natural logarithm of π‘₯ appears in our integrand. And if we differentiate this, we get one over π‘₯. So our integrand is a function multiplied by its derivative. This means we can evaluate this by using substitution. We’ll use the substitution 𝑒 is equal to the natural logarithm of π‘₯. Differentiating both sides with respect to π‘₯, we get d𝑒 by dπ‘₯ is equal to one over π‘₯. And we know d𝑒 by dπ‘₯ is not a fraction. However, when using integration by substitution, we can treat it a little bit like a fraction. This gives us the equivalent statement d𝑒 is equal to one over π‘₯ dπ‘₯.

We’re almost ready to apply our substitution. But remember, this is a definite integral. So we need to calculate the new limits of our integration. Let’s start with the new lower limit of our integral. Our old lower limit was when π‘₯ is equal to three. So we substitute this into our expression for 𝑒, and we get 𝑒 is equal to the natural logarithm of three. And we’ll do the same to find the new upper limit of our integral. We get the natural logarithm of 𝑑. So we’re now ready to evaluate this integral by using substitution. We’ll start by rewriting the limits of our integration. We get the lower limit is the natural logarithm of three and the upper limit is the natural logarithm of 𝑑.

Next, remember, the natural logarithm of π‘₯ is equal to 𝑒. And we know that one over π‘₯ times dπ‘₯ is equal to d𝑒. So we’ve simplified this expression to be the limit as 𝑑 approaches ∞ of the integral from the natural logarithm of three to the natural logarithm of 𝑑 of 𝑒 with respect to 𝑒. And we can then just evaluate this integral by using the power rule for integration. We get the limit as 𝑑 approaches ∞ of one-half 𝑒 squared evaluated at the limits of integration, the natural logarithm of three and the natural logarithm of 𝑑. And we’ll simplify this expression slightly. We’ll take the constant factor of one-half outside of our limit.

Now we just evaluate this at the limits of integration. We get one-half times the limit as 𝑑 approaches ∞ of the natural logarithm of 𝑑 all squared minus the natural logarithm of three all squared. And we can now just evaluate this limit as it is. We see 𝑑 is approaching ∞. We know the natural logarithm of three all squared is a constant. It won’t change as the value of 𝑑 changes. However, as 𝑑 is approaching ∞, the natural logarithm of 𝑑 all squared is growing without bound. And this in fact tells us our entire limit is growing without bound. So our limit evaluates to give us ∞.

So this limit is divergent. And remember, this limit being divergent means that our integral from three to ∞ of the natural logarithm of π‘₯ divided by π‘₯ with respect to π‘₯ is also divergent. So we’re in the second part of our integral test. This means we know the sum from 𝑛 equals three to ∞ of π‘Ž 𝑛 is also divergent. And we already explained this is equivalent to saying the series given to us in the question is divergent. Therefore, by using the integral test, we were able to show the sum from 𝑛 equals one to ∞ of the natural logarithm of 𝑛 divided by 𝑛 is divergent.

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